Problem: Evaluate $\int\cos^2x\tan^3x\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\cos^2(x)}{2} - \ln|\cos x | + C$ (Choice B) B $\dfrac{\tan^2(x)}{2} - \ln|\tan x | + C$ (Choice C) C $\dfrac{\sin^2(x)}{2} - \ln|\sin x | + C$ (Choice D) D $\dfrac{\sin^3(x)}{3} - \ln|\sin x | + C$
Explanation: First write $\tan x$ as $\dfrac{\sin x}{\cos x}$. $\int\cos^2x\tan^3x\,dx=\int \cos^2x \dfrac{\sin^3x}{\cos^3x} dx= \int \dfrac{\sin^3x}{\cos x} dx$ Next replace a factor of $\sin^2x$ with $1-\cos^2x$. $ \int \dfrac{\sin^3x}{\cos x} dx= \int \dfrac{\sin x}{\cos x} \sin^2x dx = \int \dfrac{\sin x }{\cos x } \Big( 1 - \cos^2 x\Big) dx$ Next we use a $u$ -substitution with $u=\cos x$ and $du=-\sin x \,dx$. $\begin{aligned} &\phantom{=} \int \dfrac{\sin x }{\cos x } \Big( 1 - \cos^2 x\Big) dx \\\\ &=\int \dfrac{\cos^2x-1}{\cos x} (-\sin x) dx \\\\ &=\int \left( \dfrac{u^2 - 1}{u} \right) du \\\\ &= \int \Big(u-\dfrac1u\Big)\,du \end{aligned}$ Finally, we do the integration and replace all appearances of $u$ with $\cos x$. $\begin{aligned} &\phantom{=}\int \Big(u-\dfrac1u\Big)\,du \\\\ &= \dfrac{u^2}2-\ln|u|+C \\\\ &=\dfrac{\cos^2x}{2} - \ln|\cos x | + C \end{aligned}$